[telmo_d]

I have too create an output similar to this for input 5.

15
24
33
42
51

I have the code below but its just printing me this:
5
1
2
3
4
5

I don't understand why. Why is that?

Code

#include <stdio.h>
#include <stdlib.h>

int main()

{

    int x, y, z;
    printf("Introduce a number: \n");
    scanf("%d", &z);
    for (x=1;x<=z;x++)
        {
            printf("%d\n", x);
            for (int y=z;y<=0;y--)
        {
                printf("   %d\n", y);
                
                
        }
        }

[TheComet]

A much easier solution:

PHP Code


<?php 
#include <stdio.h>

int main()
{
    int x, y;

    printf("Introduce a number: \n");
    scanf(" %d", &x);

    y = 1;
    while(x)
    {
      printf("%d%d\n", y++, x--);
    }

    return 0;
} 


[telmo_d]

A much easier solution:

PHP Code


<?php 
#include <stdio.h>

int main()
{
    int x, y;

    printf("Introduce a number: \n");
    scanf(" %d", &x);

    y = 1;
    while(x)
    {
      printf("%d%d\n", y++, x--);
    }

    return 0;
} 


That's a nice solution. I have to really think all my options next time but i've been reading nested for loops and decided to aplly that, clearly i have trouble with them.
Could you change the code to make it work? I can't see what's wrng.

[TheComet]

i've been reading nested for loops and decided to aplly that

It's not a nested loop problem. You cannot solve this using nested loops, because the output of nested loops would be n^2, but you only have n outputs.

[Sights101]

You could do a two variable for loop if its easier for you to visualize...

Code

#include <iostream>
using namespace std;

int main() {
    int x;
    cout << "Input a number" << endl;
    cin >> x;
    for (int i = 1, j = x; i <= x; ++i, --j) {
      cout << i << j << endl;
    }
}

[0xDEAD10CC]

I don't know why you guys are making this so difficult... Anyone notice that the values are increments of 9? lol! 🤝

Here, a good mathematical way to solve this:

Code

int input = 9;
for (int i = 0, x = input + 10; i < input; ++i, x += 9)
{
  printf("%d\n", x);
}

Works for 1 >= input <= 9.. I wouldn't know what kind of output you'd expect if input was 0... Perhaps just "0"? or "00"? (Your example starts at 1 however)... Or if input was greater than 9.